Simplify Boolean Algebra expression $A\overline C + A\overline B + \overline CB$

Question

Can anyone explain to me how

$$A\overline C + A\overline B + \overline CB$$

simplifies down to $A\overline B +B \overline C$ ?

Answer 1

Notice that $$A\bar C=A\bar C(B+\bar B)=A\bar CB+A\bar C\bar B$$ Therefore $$\begin{align}A\bar C+A\bar B+\bar CB&=A\bar CB+A\bar C\bar B+A\bar B+\bar CB\\&=A\bar B(\bar C+1)+\bar CB(A+1)\\&=A\bar B+\bar CB\end{align}$$

Answer 2

I agree with @Fabio Somenzi that there are more efficient ways that using Karnaugh maps/tables. Nevertheless, in the spirit of SE, where one tries to learn how to solve issues with different methods, and for people that don't know this method, here is a proof using these Karnaugh maps:

$$\begin{array}{|c|c|c|c|} \hline &\color{red}{B}&\color{red}{B}&\color{red}{\bar B}&\color{red}{\bar B}\\ \hline \color{red}{A}&A B C &A B \bar C&\bar A \bar B \bar C& A \bar B C\\ \hline \color{red}{\bar A}&\bar A B C&\bar A B \bar C&\bar A \bar B \bar C&\bar A \bar B C\\ \hline &\color{red}{C}&\color{red}{\bar C}&\color{red}{\bar C}&\color{red}{C}\\ \hline \end{array}$$

One has only to check that the two expressions have the same Karnaugh table which is:

$$\begin{array}{|c|c|c|c|} \hline &\color{red}{B}&\color{red}{B}&\color{red}{\bar B}&\color{red}{\bar B}\\ \hline \color{red}{A}& 0&1&1&1 \\ \hline \color{red}{\bar A}&0&1&0&0\\ \hline &\color{red}{C}&\color{red}{\bar C}&\color{red}{\bar C}&\color{red}{C}\\ \hline \end{array}$$

Remark: "Algebraic derivations" like the one given by @PirateBay are almost impossible to find without using this kind of table or similar methods.